What is the integral of 1/(2cos^2(3x)) ?

The integral of 1/(2cos^2(3x)) is 1/6 tan(3x)+C

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integral of 1/(2cos^2(3x))

\int \frac{1}{2 cos ^{2} ( 3 x )} d x

\frac{1}{6} tan (3 x) + C

Solution steps

\int \frac{1}{2 cos ^{2} ( 3 x )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{cos ^{2} ( 3 x )} d x

Rewrite using trig identities

= \frac{1}{2} \cdot \int sec^{2} (3 x) d x

Apply u-substitution

= \frac{1}{2} \cdot \int sec^{2} (u) \frac{1}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= \frac{1}{2} \cdot \frac{1}{3} tan (u)

Substitute back

u = 3 x

= \frac{1}{2} \cdot \frac{1}{3} tan (3 x)

Simplify

\frac{1}{2} \cdot \frac{1}{3} tan (3 x) : \frac{1}{6} tan (3 x)

= \frac{1}{6} tan (3 x)

Add a constant to the solution

= \frac{1}{6} tan (3 x) + C

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What is the integral of 1/(2cos^2(3x)) ?
The integral of 1/(2cos^2(3x)) is 1/6 tan(3x)+C